Solve the ODE $y'^{2}+y^2=1$ I can figure out solutions $y=\sin x$ or $\cos x$ and trivial solution $y=1$. But how to get all solutions?
 A: Hint
If you're looking for the solutions around $x=0$ and supposes that $\vert y(0) \vert \lt 1$ then the ODE is equivalent to
$$\frac{y^\prime(x)}{\sqrt{1 - y^2(x)}} = \pm 1$$ which can be integrated as
$$\arcsin\left(y(x) \right) = \pm x + C$$ and leads to
$$y(x) = \sin\left(\pm x +C\right).$$
While if $\vert y(0) \vert \gt 1$ there is no solution and the solution is constant when $\vert y(0) \vert =1$.
A: Let $I$ an interval in $ \mathbb R$ and let $y:I \to \mathbb R$ be a solution of the ODE.
Furthermore we assume that $y'(x) \ne 0$ for all $x \in I$. This gives $|y(x)|<1$ for all $x \in I.$ From $y'(x)= \pm \sqrt{1-y(x)^2}$, we see that $y'$ is differentiable on $I$ and
$$2y'(x)y''(x)+2y'(x)y(x)=0$$
for all $x \in I.$ Hence
$$ y''+y=0.$$
The last ODE has the general solution $y(x)=a \cos x+ b \sin x.$ It is now easy to see that $a^2+b^2=1.$
Conclusion:
A: It is $$\int \frac{dy}{\sqrt{1-y^2}}=\pm \int dx$$
$$\implies \sin^{-1} x=\pm x+C \implies y(x)=\pm \sin (x\pm C)$$
So the general solution can also be written as $y=\pm [\sin x \cos C\pm \cos x \sin C]$ If $C=0$ Solution is $y=\pm \sin x$, if $C=\pi/2$ the solution is $y=\pm \cos x$.$y(x)=1$ is singular (essential) solution of this ODE which is free of constant.
A: $$y'^2=1-y^2.$$
If $y^2=1$, we get $y'=0$ and we have two constant solutions $y=\pm1$.
Otherwise,
$$\frac{y'^2}{1-y^2}=1$$ or $$\frac{y'}{\sqrt{1-y^2}}=\pm1^*.$$
Then by integration,
$$\arcsin(y)-\arcsin(y_0)=\pm x$$ and
$$y=\sin(\arcsin(y_0)\pm x).$$

$^*$Note that we cannot switch from $+1$ to $-1$ at some arbitrary $x$, as this would introduce a discontinuity in the derivative. Hence the sign remains constant for all $x$.
A: Failed attempt:
As the $(y,y')$ trajectory is a circle, we can write the solution as $$y(x)=\sin(u(x)),y'(x)=\cos(u(x)).$$ This requires
$$y'(x)=\cos(u(x))=u'(x)\cos(u(x))$$ and $u'(x)=1$ or $\cos(u(x))=0$.
The first case yields $u(x)=x+c$, or $y(x)=\sin(x+c)$, and the second is $y(x)=\pm1$.
Unfortunately, this method does not yield the most general solution because it does not cover the case
$$y(x)=\cos(u(x)),y'(x)=\sin(u(x))$$ that results in $u'(x)=-1$.
